Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $p = \dfrac{-2n + 16}{-3n^2 + 9n + 120} \div \dfrac{n - 8}{n^2 + 14n + 45} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{-2n + 16}{-3n^2 + 9n + 120} \times \dfrac{n^2 + 14n + 45}{n - 8} $ First factor out any common factors. $p = \dfrac{-2(n - 8)}{-3(n^2 - 3n - 40)} \times \dfrac{n^2 + 14n + 45}{n - 8} $ Then factor the quadratic expressions. $p = \dfrac {-2(n - 8)} {-3(n + 5)(n - 8)} \times \dfrac {(n + 5)(n + 9)} {n - 8} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac {-2(n - 8) \times (n + 5)(n + 9) } { -3(n + 5)(n - 8) \times (n - 8)} $ $p = \dfrac {-2(n + 5)(n + 9)(n - 8)} {-3(n + 5)(n - 8)(n - 8)} $ Notice that $(n + 5)$ and $(n - 8)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {-2\cancel{(n + 5)}(n + 9)(n - 8)} {-3\cancel{(n + 5)}(n - 8)(n - 8)} $ We are dividing by $n + 5$ , so $n + 5 \neq 0$ Therefore, $n \neq -5$ $p = \dfrac {-2\cancel{(n + 5)}(n + 9)\cancel{(n - 8)}} {-3\cancel{(n + 5)}\cancel{(n - 8)}(n - 8)} $ We are dividing by $n - 8$ , so $n - 8 \neq 0$ Therefore, $n \neq 8$ $p = \dfrac {-2(n + 9)} {-3(n - 8)} $ $ p = \dfrac{2(n + 9)}{3(n - 8)}; n \neq -5; n \neq 8 $